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2006 Q7(c)(iii) 2006 Q7(c)(iii)


Posted 26 April 2008 - 03:15 PM

can anyone help with this plz....
if 1.64kg of ethanol (relative formula mass = 46) is produced from 10.0kg of ethene (relative formula mass = 28),calculate the percentage yield of ethanol.


Posted 11 May 2008 - 06:55 PM

First you should write out the equation and balance it in order to find the mole ratio of reactants and products.

C_2H_4 + H_2O \rightarrow C_2H_5OH

In this case the ratio is 1 : 1 and the equation is already balanced. Now you have to find the number of mols of reactant used, ie. the number of mols of ethene.

Mols = \frac{Actual Mass}{Formula Mass} = \frac{10000}{28} = 357.143

In theory, since the ratio is 1 : 1, 357.143 mols of ethene will produce 357.143 mols of ethanol when hydrated. In order to calculate the actual percentage yield you must find the theoretical mass of ethanol produced.

By rearranging the above equation:

\begin{align*}Actual Mass &= Mols \times Formula Mass \\
&= 357.143 \times 46 \\
&= 16428.571g \\
&= 16.43kg\end{align*}

Now you simply use the equation to calculate percentage yield.

\begin{align*}Percentage Yield &= \frac{Actual Mass}{Theoretical Mass} \times 100\\
&= \frac{1.64}{16.43} \times 100 \\
&= 9.98\end{align*}

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